oliviayychengwh

Answered

2022-01-02

How to prove this inequality in interval: $(0,\frac{\pi}{2})\text{}\text{}2\mathrm{cos}x+{\mathrm{sec}}^{2}x-3\ge 0$

Answer & Explanation

Thomas Nickerson

Expert

2022-01-03Added 32 answers

Restrict the domain to some meaningful interval such as $x\in (-\frac{\pi}{2},\frac{\pi}{2})$ so that $\mathrm{cos}x>0$ . In that case, using the AM-GM inequality you have:

$LHS=\mathrm{cos}x+\mathrm{cos}x+{\mathrm{sec}}^{2}x-3\ge 3\sqrt[3]{\mathrm{cos}x\cdot \mathrm{cos}x\cdot {\mathrm{sec}}^{2}x}-3=3-3=0=RHS$

Papilys3q

Expert

2022-01-04Added 34 answers

We have

$2\mathrm{cos}x+{\mathrm{sec}}^{2}x-3=2(\mathrm{cos}x-1)+(\frac{1}{{\mathrm{cos}}^{2}x}-1)=\frac{(2\mathrm{cos}x+1){(\mathrm{cos}x-1)}^{2}}{{\mathrm{cos}}^{2}x}$

Thefore the inequality does not hold.

Thefore the inequality does not hold.

Vasquez

Expert

2022-01-09Added 457 answers

By Fermat's theorem, we examine zeroes of derivative of the function

The solution is

Thus

Since the interval is added, we can prove the claim is true. For x=0, f(x)=0. And when

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